1.8.8 Orders of Magnitude and Estimating

Precise calculations are appropriate when we have precise data, but there are many times when a very rough estimate is sufficient. Suppose you see a rock fall off a cliff and would like to know how fast it was going when it hit the ground. By doing a mental comparison with the speeds of familiar objects, such as cars and bicycles, you might judge that the rock was traveling at “about” $20 mph$ .

This is a one-significant-figure estimate. With some luck, you can distinguish $20 mph$ from either $10 mph$ or $30 mph,$ but you certainly cannot distinguish $20 mph$ from $21 mph$ . A one-significant-figure estimate or calculation, such as this, is called an order-of-magnitude estimate. An order-of-magnitude estimate is indicated by the symbol $\sim$ , which indicates even less precision than the “approximately equal” symbol $\approx$ . You would say that the speed of the rock is $v \sim 20 mph$ .

A useful skill is to make reliable estimates on the basis of known information, simple reasoning, and common sense. This is a skill that is acquired by practice. Many chapters in this book will have homework problems that ask you to make order-of-magnitude estimates. The following example is a typical estimation problem.

TABLES 1.6 and 1.7 have information that will be useful for doing estimates.

TABLE 1.6 Some approximate lengths

	Length (m)
Altitude of jet planes	10,000
Distance across campus	1000
Length of a football field	100
Length of a classroom	10
Length of your arm	1
Width of a textbook	0.1
Length of a fingernail	0.01

TABLE 1.7 Some approximate masses

	Mass (kg)
Small car	1000
Large human	100
Medium-size dog	10
Science textbook	1
Apple	0.1
Pencil	0.01
Raisin	0.001

EXAMPLE 1.11 Estimating a sprinter’s speed

Estimate the speed with which an Olympic sprinter crosses the finish line of the $100 m$ dash.

SOLVE We do need one piece of information, but it is a widely known piece of sports trivia. That is, world-class sprinters run the $100 m$ dash in about $10 s$ . Their average speed is $v_{avg} \approx (100 m) / (10 s) \approx 10 m / s$ . But that’s only average. They go slower than average at the beginning, and they cross the finish line at a speed faster than average. How much faster? Twice as fast, $20 m / s$ , would be $\approx 40 mph$ . Sprinters don’t seem like they’re running as fast as a $40 mph$ car, so this probably is too fast. Let’s estimate that their final speed is 50% faster than the average. Thus they cross the finish line at $v \sim 15 m / s$ .